Department of Physics
University of Maryland
Physics 121 Fall 2002
Homework Assignment # 8
Problem Solutions
5)For this problem we recall the result
α =
∆ω
∆t
as well we know 1 rev = 2 π rad and 1 min = 60 s. It follows that
rev
2 π rad
rad
= 210
= 21.9
min
60 s
s
2 π rad
rad
rev
= 480
= 50.3
ω0 = 480
min
60 s
s
rad
rad
→ ∆ ω = [ 21.9 − 50.3 ]
= − 28.4
s
s
∆ t = 74 (60) s = 4440 s
ωf = 210
So that finally
α =
− 28.4 rad
rad
s
= − 6.4 × 10−3 2
4440 s
s
The magnitude is just 6.4 × 10−3
rad
.
s2
12) At t seconds, the angular positions for the two people are respectively
given by
θ1 = ω1 t , θ2 = ω2 t
For them to meet the sum of their two angular positions must equal
2 π rad and this leads to the equation
ω1 t + ω2 t = 2 π rad
t =
t =
2 π rad
ω1 + ω2
1.7 ×
2 π rad
+ 3.4 × 10−3
10−3 rad
s
rad
s
14) In class we saw why the formula
s = rθ
→ θ =
s
r
= 1200 s
is valid. This problem requires that we make use of this twice. We are
given the numerical data
sm = 3.48 × 106 m ,
rm = 3.85 × 108 m
sS = 1.39 × 109 m ,
rS = 1.50 × 1011 m
so that we can calculate the angle subtended by each as we see them
in the sky is given by
3.48 × 106 m
1.39 × 109 m
,
θ
=
m
3.85 × 108 m
1.50 × 1011 m
= 9.04 × 10−3 rad , θS = 9.27 × 10−3 rad
θm =
θm
It can be seen that the angle for the sun is a little bit larger than that of
the moon. This means that the moon cannot completely “block out” the
sun. The areas that the moon and sun block out are respectively given by
Am = 14 π (sm )2 , AS = 14 π (sS )2
Am = 14 π (rS θm )2 , AS = 14 π (rS θS )2
so that the ratio of the areas is given by
θ 2
9.04 × 10−3 2
Am
(rS θm )2
m
=
=
=
AS
(rS θS )2
θS
9.27 × 10−3
9.04 2
Am
=
= 0.951
AS
9.27
and the moon only blocks out 95.1% of the sun.
22) One of the equations of linear kinematics is given by
xf = x0 + v0 t + 12 a t2
and as discussed in class this can directly be used to derive
θf = θ0 + ω0 t + 12 α t2
so that upon using the numerical data this becomes
θf = 0 + (74.5
rad
rad
) (4.5 s) + 12 (−6.7 2 ) (4.5 s)2
s
s
θf = 267 rad
29) One revolution is 2 π rad so we use
ω =
∆θ
2 π rad
rad
=
= 0.33
∆t
18.9 s
s
2
and the linear speed is related to the angular speed via v = R ω or equivalently
R =
42.6 m
v
s
=
rad = 128 m
ω
0.33 s
38) We recall that the centripetal acceleartion ac is related to the angular
speed via ac = R ω 2 so
ac, 1 = R (ω1 )2
,
ac, 2 = R (ω2 )2
and dividing the first by the second we see
ω 2
ac, 1
R (ω1 )2
(ω1 )2
1
=
=
=
2
2
ac, 2
R (ω2 )
(ω2 )
ω2
ac, 1
440 2
=
= 16
ac, 2
110
39) The centripetal and tangential acceleration are given respectively by
ac = r ω 2
, aT = r α
and these two components are at right angles to one another so the magnitude of the acceleration is given by the Pythagorean theorem
a =
q
= r
(ac )2 + (aT )2 =
q
q
(ω 2 )2 + (α)2 = r
s
= (1.5 m)
(14
(r ω 2 )2 + (r α)2
q
(ω)4 + (α)2
rad 4
rad
) + (160 2 )2
s
s
= (1.5)
q
(14)4 + (160)2 sm2
= (1.5)
q
(38416) + (25600) sm2
= (1.5)
q
(64016) sm2 = (1.5) 253 sm2 = 380 sm2
40) We need the following three equations for this problem.
ω = ω =
∆θ
∆t
,
vT = r ω
,
ac = r ω 2
The first of these can be substituted into the last two
vT = r
∆ θ
∆t
,
ac = r
∆ θ 2
∆t
The numerical data are
∆ θ = 2 π rad , ∆ t = 3.16 × 107 s , r = 1.5 × 1011 m
3
and this leads to
∆θ
2 π rad
rad
=
= 1.99 × 10−7
7
∆t
3.16 × 10 s
s
rad
vT = (1.5 × 1011 m) (1.99 × 10−7
) = 2.98 × 104 m
s
s
rad 2
ac = (1.5 × 1011 m) (1.99 × 10−7
) = 5.94 × 10−3 sm2
s
41) Newton’s second law implies
aT =
F
550 N
=
= 2.5 sm2
m
220 kg
and we can use one of the kinematical equations to find the tangential speed.
m
m
vT = V0 + aT t = (5 m
s ) + (2.5 s2 ) (2 s) = 10 s
and finally for the centripetal accelearation
ac =
2
(10 m
(vT )2
s ) = 3.1 m
=
s
r
32 m
42) The tangential speed all along the sprocket must be the same and given by
vT = r ω = (9 cm) (9.4
rad
) = 84.6 cm
s
s
after using the numerical data. At the rear we have the equations
ac = rrear (ω)2
,
ω =
vT
rrear
The second one of these can be substituted into the first to yield
ac = rrear
v 2
T
rrear
=
(vT )2
rrear
and finally we use the numerical data
2
(84.6 cm
s ) = 1.4 × 103 cm
ac =
s2
5.1 cm
4